2.) Again, using mods, show if gcd(n,100) = 1, then n^41 has the same last 2 digits as n. Explain if it is true for ALL n, that n and n^41 have the same last 2 digits.

By Euler Elegant Generalization of the Super Little Theorem of the Cool Fermat, we have that,

a^{phi(n)} = 1 (mod n)

In this case let n=100.
Thus, phi(100)=40.
Thus,
a^{40} = 1 (mod 100)
Where gcd(a,n)=100.
Then, multiplication by "a" yield,
a^{41} = a (mod 100}
Thus they have the same last two digits for they are congruent to each other.

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Consider n=2. The last two digits are 02 (if that is how you want to think of it).